KVPY Solution 25-10-10

KVPY QUESTION WITHANSWER(25-10-10)

1.

Sol.

If the angle of elevation of a cloud c loud from a point 200 metres above a lake is 30º and the angle of  Sol. depression of its reflection in the lake is 60º, then the height of the cloud (in metres) above the lake is (A) 200 (B) 300 (C*) 400 (D) 500 In ABC

Let the two spheres collide at point A, after time t, the distance covered by smaller sphere is x1 and bigger sphere is x2. Gravitational force on each sphere, x1 A x2 2R R 9R •

•

GM  5M

F=

(12R)2 F M

Acceleration of smaller sphere, a1 = G  5M

=

(12R )2

Acceleration of bigger sphere, a2 = GM

=

(12R )2

Using equation of motion , s = ut +

x1 a1 = =5 x2 a2



x1 = 5x2

--- (i)

Also, we know , x 1 + x2 = 9R --- (ii) From (i) and (ii) we get x1 = 7.5 R

x = h 3 ....(i) In BDC h  400 tan 60° = x

3 x = h + 400 ...(ii) from (i) and (ii)

Sol.

No change in moment of inertia

3.

Conjugate base of [C2H5NH3]+ is (A*) C2H5NH2

(B) C2N+H4(OH)

(C) [C2H2NH]

(D) [C2H5]+

 –

 –

h 3 . 3 = h + 400 3h h = 400  –

Sol.

400 = 200 2 h = 200 So, height of object above the lake = h + 200 = 200+200 = 400 Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to

xx

CH 2

5

Base

h=

2.

1 2 at 2

1 1 × a × t2 and x2= a t2 2 2 2

x1 = h tan 30° = x

F 5M

4.

 –

NH + H O 2

2

Acid

CH 2

Acid

(A*) phylogenetic classification (B) artificial classification

by the smaller body before collision is :

(D) numerical classification

(C) 1.5 R

(D*) 7.5 R

 –

3

Base

interrelationships of organisms is called

(C) natural classification

(B) 4.5R

NH + OH

Classification reflecting the evolutionary

gravitational force forc e only, only, then the distance covered (A) 2.5R

5

+

 –